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1966 IMO Problems/Problem 3

Revision as of 01:23, 27 September 2024 by Pf02 (talk | contribs)

Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.

Solution

We will need the following lemma to solve this problem:

$\emph{Lemma:}$ Let $MNOP$ be a regular tetrahedron, and $T$ a point inside it. Let $x_1, x_2, x_3, x_4$ be the distances from $T$ to the faces $MNO, MNP, MOP$, and $NOP$. Then, $x_1 + x_2 + x_3 + x_4$ is constant, independent of $T$.

$\emph{Proof:}$

We will compute the volume of $MNOP$ in terms of the areas of the faces and the distances from the point $T$ to the faces:

\[\textrm{Volume}(MNOP) = [MNO] \cdot x_1 \cdot \frac{1}{3} + [MNP] \cdot x_2 \cdot \frac{1}{3} + [MOP] \cdot x_3 \cdot \frac{1}{3} + [NOP] \cdot x_4 \cdot \frac{1}{3}\] \[= [MNO] \cdot \frac{(x_1 + x_2 + x_3 + x_4)}{3}\]

because the areas of the four triangles are equal. ($[ABC]$ stands for the area of $\triangle ABC$.) Then

\[\frac{3\cdot\textrm{Volume}(MNOP)}{[MNO]} = x_1 + x_2 + x_3 + x_4.\]

This value is constant, so the proof of the lemma is complete.

$\emph{Proof of problem statement:}$

Let our tetrahedron be $ABCD$, and the center of its circumscribed sphere be $O$. Construct a new regular tetrahedron, $WXYZ$, such that the centers of the faces of this tetrahedron are at $A$, $B$, $C$, and $D$.

For any point $P$ in $ABCD$,

\[OA + OB + OC + OD = \sum \textrm{Distances from }O\textrm{ to faces of }WXYZ\] \[= \sum \textrm{Distances from }P\textrm{ to faces of }WXYZ \leq PA + PB + PC + PD,\]

with equality only occurring when $AP$, $BP$, $CP$, and $DP$ are perpendicular to the faces of $WXYZ$, meaning that $P = O$. This completes the proof. $\square$

~mathboy100


Remarks (added by pf02, September 2024)

1. The text of the Lemma needed a little improvement, which I did.

2. The Solution above is not complete. It considered only points $P$ inside the tetrahedron, but the problem specifically said "any other point in space".

3. I will give another solution below, in which I will also fill in the gap of the solution above, mentioned in the preceding paragraph.


Solution 2

We will first prove the problem in the 2-dimensional case. We do this to convey the idea of the proof, and because we will use this in one spot in proving the 3-dimensional case. So let us prove that:

The sum of the distances of the vertices of an equilateral triangle $\triangle ABC$ from the center $O$ of its circumscribed circle is less than the sum of the distances of these vertices from any other point $P$ in the plane.

We will do the proof in three steps:

1. We will show that if $P$ is in one of the exterior regions, then there is a point $P_1$ on the boundary of the triangle (a vertex, or on a side), such that $PA + PB + PC > P_1A + P_1B + P_1C$.

2. Then we will show that if $P$ is on the boundary, then $PA + PB + PC > OA + OB + OC$.

3. For the final step, we will show that if $P$ is a point of minimum for $PA + PB + PC$ inside the triangle, then the extensions of $PA, PB, PC$ are perpendicular to the opposite sides $BC, AC, AB$. This implies that $P = O$.

Proof of 1: If the point $P$ is outside the triangle, it can be in one of six regions as seen in the pictures below.

Prob 1966 3 fig1.png

If $P$ is in a region delimited by extensions of two sides of the triangle, as in the picture on the left, we notice that by taking $P_1 = A$, $PA + PB + PC > P_1A + P_1B + P_1C$ (because $P_1A = 0$ and $P_1B < PB$ as sides in an obtuse triangles, and similarly $P_1C < PC$).

If $P$ is in a region delimited by a segment which is a side of the triangle and by the extensions of two sides, as in the picture on the right, take $P_1 =$ the foot of the perpendicular from $P$ to $AB$. Then $PA + PB + PC > P_1A + P_1B + P_1C$ (because the triangle $\triangle PP_1C$ is obtuse, and because the triangles $\triangle PP_1A, \triangle PP_1B$ are right triangles).

Proof of 2: Now assume that $P_1 = A$. A direct, simple computation shows that $P_1A + P_1B + P_1C > OA + OB + OC$ (indeed, if we take the side of the triangle to be $1$, then $P_1A + P_1B + P_1C = 2$, and $OA + OB + OC = 3 \cdot \frac{\sqrt{3}}{3} = \sqrt{3}$).

Now assume that $P_1$ is on $AB$. If $P_1$ is not the midpoint of $AB$, let $P_2$ be the midpoint. Then $P_1A + P_1B + P_1C > P_2A + P_2B + P_2C$ (because $P_1A + P_1B = P_2A + P_2B = AB$ and $P_1C > P_2C$). A direct, simple computation shows that $P_2A + P_2B + P_2C > OA + OB + OC$ (indeed, if we take the side of the triangle to be $1$, $P_2A + P_2B + P_2C = 1 + \frac{\sqrt{3}}{2}$ and $OA + OB + OC = \sqrt{3}$).

Proof of 3: Assume that $P$ is inside the triangle $\triangle ABC$. In this case, we make a proof by contradiction. We will show that if $P$ is a point where $PA + PB + PC$ is minimum, then the extensions of $PA, PB, PC$ are perpendicular to the opposite sides $BC, AC, AB$. (This statement implies that $P = O$.) If this were not true, at least one of $PA \perp BC, PB \perp AC, PC \perp AB$ would be false. We can assume that $PC$ is not perpendicular to $AB$. Then draw the ellipse with focal points $A, B$ which goes through $P$.

Prob 1966 3 fig2.png

Now consider the point $P_1$ on the ellipse such that $CP_1 \perp AB$. Because of the properties of the ellipse, $CP_1 < CP$, and because of the definition of the ellipse $PA + PB = P_1A + P_1B$. We conclude that $PA + PB + PC > P_1A + P_1B + P_1C$, which contradicts the assumption that $P$ was such that $PA + PB + PC$ was minimum.

This proves the 2-dimensional case.

One note: a very picky reader might object that the proof used that a minimum of $PA + PB + PC$ exists, and is achieved at a point $P$ inside the triangle. This can be justified simply by noting that $PA + PB + PC > 0$ and quoting the theorem from calculus (or is it topology?) which says that a continuous function on a closed, bounded set has a minimum, and there is a point where the minimum is achieved. Because of the arguments in the proof, this point can not be on the boundary of the triangle, so it is inside.

Now we will give the proof in the 3-dimensional case. We will do the proof in three steps. It is extremely similar to the proof in the 2-dimensional case, we just need to go from 2D to 3D, so I will skip some details.

1. We will show that if $P$ is in one of the exterior regions, then there is a point $P_!$ on the boundary of the tetrahedron (a vertex, or on a edge, or on a side, such that $PA + PB + PC + PD > P_1A + P_1B + P_1C + P_1D$.

2. Then we will show that if $P$ is on the boundary, then $PA + PB + PC + PD > OA + OB + OC + OD$.

3. For the final step, consider the plane going through the edge $CD$ perpendicular to the edge $AB$, the plane going through $AB$ perpendicular to $CD$, the plane going through $CA$ perpendicular to $BD$, etc. There are six such planes, and they all contain $O$, the center of the circumscribed sphere. We will show that if $P$ is a point of minimum for $PA + PB + PC + PD$ inside the tetrahedron, then $P$ is in each of the six planes described above. This implies that $P = O$.

Proof of 1:



TO BE CONTINUED. SAVING MID WAY SO I DON'T LOSE WORK DONE SO FAR.

(Solution by pf02, September 2024)


See Also

1966 IMO (Problems) • Resources
Preceded by
Problem 2 		1 • 2 • 3 • 4 • 5 • 6 Followed by
Problem 4
All IMO Problems and Solutions
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